How do you integrate $sin (ln (4 x + 5))$ $sin(ln(4x+5))$ ?

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How do you integrate $sin (ln (4 x + 5))$ $sin(ln(4x+5))$ ?

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Answer 1 · 2018-05-27T17:53:25

$I = \frac{4 x + 5}{8} [sin (ln (4 x + 5)) - cos (ln (4 x + 5))] + C$ $I=(4x+5)/8[sin(ln(4x+5))-cos(ln(4x+5))]+C$

Explanation:

Here,

$I = \int sin (ln (4 x + 5)) d x$ $I=intsin(ln(4x+5))dx$

Subst. $ln (4 x + 5) = u \Rightarrow 4 x + 5 = e^{u}$ $color(violet)(ln(4x+5)=u=>4x+5=e^u$

$\Rightarrow 4 d x = e^{u} d u \Rightarrow d x = \frac{1}{4} e^{u} d u$ $=>4dx=e^udu=>dx=1/4e^udu$

So,

$I = \int sin u \times \frac{1}{4} e^{u} d u$ $I=intsinuxx1/4e^udu$

$=>4I=intsinue^udu...to(A)$

$"Using "color(blue)"Integration by Parts :"$

$color(red)(intf(x)g'(x)dx=f(x)g(x)-intf'(x)g(x)dx$

$=>4I=sinuinte^udu-int(cosuinte^udu)du$

$=>4I=sinu*e^u-intcosue^udu$

$=>4I=sinue^u-I_1...to(B)$

Where, $I_1=intcosue^udu$

Again $"using "color(blue)"Integration by Parts :"$

$:.I_1=cosuinte^udu-int(-sinuinte^udu)du$

$:.I_1=cosue^u+intsinue^udu+c'$

$:.I_1=cosue^u+4I+c'...to$ from $(A)$

From $(B)$ we get,

$:.4I=sinue^u-{cosue^u+4I}+c,where,c=-c'$

$:.4I=sinue^u-cosue^u-4I+c$

$4I+4I=sinue^u-cosue^u+c$

$8I=e^u(sinu-cosu)+c$

$I=1/8e^u(sinu-cosu)+C,where, C=c/8$

Subst, back , $color(violet)(u=ln(4x+5) and e^u=4x+5$

$I=1/8(4x+5)[sin(ln(4x+5))-cos(ln(4x+5))]+C$

$I=(4x+5)/8[sin(ln(4x+5))-cos(ln(4x+5))]+C$

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How do you integrate $sin (ln (4 x + 5))$ $sin(ln(4x+5))$ ?

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